MAT244--2018F > Term Test 1

TT1 Problem 3 (morning)

(1/1)

**Victor Ivrii**:

(a) Find the general solution for equation

\begin{equation*}

y''-5y'+4y=-12e^t + 20 e^{-t}.

\end{equation*}

(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

**Zhihao Zuo**:

I will type the solution soon!

**Zixuan Wang**:

Since the first post did not type out the solution, I will type mine here.

Thank you Shengying, I fixed my problem here.

First, we need to find the homogeneous solution

$r^2-5r+4=0$

$(r-4)(r-1)=0$

$r_1=4,r_2=1$

$y_c(t) = c_1e^{4t}+c_2e^{t}$

Secondly, we need to find particular solution of $y" - 5y' +4y = -12e^t$

$y_p(t) =Ate^t, y'_p(t) = Ae^t +Ate^t, y"_p(t) = 2Ae^t+Ate^t$

plug in back to the equation we get A=4

Thus, $y_p(t) =4te^t$

Thirdly, we need to find particular solution of $y" - 5y' +4y = 20e^{-t}$

$y_p(t) =Bte^{-t}, y'_p(t) = -Be^{-t}, y"_p(t) = Be^{-t}$

plug in back to the equation we get B=2

Thus, $y_p(t) =2e^{-t}$

Therefore, $y(t) = c_1e^4t +c_2e^t+4te^t+2e^{-t}$

(b) $y(0) = 0$ so that $c_1+c_2 = -2$

$y'(0) =0$ so that $4c_1+c_2 = -2$

Therefore $c_1=0, c_2 = -2$ and $y(t) = -2e^t +4te^t+2e^{-t}$

**Shengying Yang**:

Zixuan, I think you made a mistake.

Plugging in $y(0)=0$ , you should get $0=C_1+C_2+2=0$. Therefore, $C_1=1, C_2=-2$

$$∴y=-2e^t+4te^t+2e^{-t}$$

**Victor Ivrii**:

Everybody was right: Zhihao solved correctly, Zixua typed (you need to learn a better way to type it) and Shengying found a mistype in the solution (not the answer)

Navigation

[0] Message Index

Go to full version